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Amiraneli [1.4K]
2 years ago
4

A bookshelf is 2m long, with supports at its end (p and q). A book weighing 10N is placed in the middle of the shelf. what are t

he upward forces acting on p and q. if the book is moved 50 cm from q, what are the forces at p and q now?
Physics
2 answers:
11Alexandr11 [23.1K]2 years ago
8 0

Answer:

Fq = 2.5N

Fp = 7.5N

Explanation:

Hello! Let's solve this!

When the book is in the middle, the force in p (Fp) and the force in q (Fq) are equal, and half of the total force

Fp = Fq = 5N

When the book moves 0.5m from q we have a balance

Fq * 2m = 10N * 0.5m

We cleared Fq

Fq = (10N * 0.5m) / 2m

Fq = 2.5N

Fp = 10N-Fq = 10N-2.5N = 7.5N

Conclusion

Fq = 2.5N

Fp = 7.5N

Aliun [14]2 years ago
7 0
When the book is placed in the middle, the forces acting on p and q is 5N. When the book is moved 50 cm from q, the forces at p and q can be solved by doing a moment balance
With p as the pivot
Fq (2 m) = 10 N (0.5 m)
Fq = 2.5 N
and
Fp = 10 N - 2.5 N = 7.5 N 
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Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

F=ma

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0
2 years ago
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A friend of yours who takes her astronomy class very seriously challenges you to a contest to find the thinnest crescent moon yo
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after the sun sets or just as it is setting

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2 years ago
If a metallic wire of cross sectional area 3.0 ´ 10-6 m2 carries a current of 6.0 A and has a mobile charge density of 4.24 ´ 10
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Answer:

The drift velocity is v_d=2.9\times 10^{-4}\ m/s.

Explanation:

Given :

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Mobile charge density , n=4.24\times 10^{28} \ carriers/m^3.

Charge value , e=1.6\times 10^{-19}\ C.

We need to find drift velocity , v_d.

Now, we know :

I=neAv_d

Therefore, v_d=\dfrac{I}{neA}

Putting all given values in above equation we get,

v_d=\dfrac{6}{4.24\times 10^{28}\times 1.6\times 10^{-19} \times 3 \times 10^{-6}}

v_d=2.9\times 10^{-4}\ m/s.

Hence, this is the required solution.

8 0
2 years ago
A flat uniform circular disk (radius = 2.00 m, mass = 1.00
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Answer:

The resulting angular speed of the disk is 0.5 rad/s

Explanation:

Step 1: Data given

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Mass of the circular disk = 1.00

Mass op the person = 40.0 kg

Distance from the axis = 1.25 m

tangential speed = 2.00 m/s

Step 2:  

There is no external torque acting on the system so we can apply the law of conservation of angular  momentum In this case the momentum is conserved.

Angular momentum of the man = Iω

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  ⇒ I = 40 *1.25² = 62.5

⇒ with ω = Angular velocity of the man

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  ⇒The time to describe this circle t = 2πr/ v

  ⇒ in 1 revolution the angle θ = 2π radians

       This angle is subtended in time t = 2πr/ v

    ⇒ The angular speed = ω = θ/t = 2π ( v/ 2πr) = v/r = 2/1.25 = 1.6 rad/s

⇒ The angular momentum of man = I*ω = 62.5 * 1.6 = 100

Since the angular momentum is conserved, before and after the man starts running we have :

Angular momentum of disk = angular momentum of the man

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or I(disk)*ω(disk) = 100

I(disk) = M(disk)*R ²/2 = 100*2*2 / 2 = 200

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The resulting angular speed of the disk is 0.5 rad/s

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2 years ago
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Answer:

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Explanation:

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Let the acceleration be a m/s².

Now, net force acting on the box towards east is given as:

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From Newton's second law of motion,

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6 0
2 years ago
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